Re: Bring Math Arguments against this FERMAT LAST THEOREM PROOF

> On 28 Aug 2005 05:41:14 -0700,
> matt271829-news@xxxxxxxxxxx wrote:
>
> >quasi wrote:
> >> On 25 Aug 2005 14:52:20 -0700,
> matt271829-news@xxxxxxxxxxx wrote:
> >>
> >> >
> >> >george ghiata wrote:
> >> >> > OBSERVATION:
> >> >> > X,Y ,Z relative Prime numbers
> >> >> > Let's say that X^3+Y^3=Z^3
> >> >> > Let"s take:
> >> >> > X+Y=W
> >> >> > and
> >> >> > R=X^2-X*Y+Y2=W^2-3*W*Y+3*Y^2=W^2-3*W*X+3*X^2
> >> >
> >> >You mean R=X^2-X*Y+Y^2, but yeah, OK.
> >> >
> >> >>
> >> >> > When Z is not divisible by 3 we see that
> (X+Y)=W
> >> >> > and R do not have any common divisor .
> >> >
> >
> >[snip verbiage]
> >
> >>
> >> Here is a proof of that part:
> >>
> >> Assume:
> >>
> >> (1) X,Y,Z are relatively prime
> >> (2) X^3 + Y^3 = Z^3
> >> (3) Z is not a multiple of 3
> >>
> >> Then R,W are relatively prime, where R=X^2-X*Y+Y^2
> and W=X+Y.
> >>
> >> proof:
> >>
> >> Suppose instead that (R,W)>1. Let p be a common
> prime factor of R,W.
> >>
> >> p|W and W*R=Z^3 => p|Z => p .ne. 3 [since Z is not
> a multiple of 3]
> >>
> >> d|W^2-R => d|3*X*Y => p|X*Y [since p .ne. 3] =>
> p|X or p|Y.
> >
> >I assume you mean p|W^2-R => p|3*X*Y => p|X*Y
> >
> >>
> >> Without loss of generality, assume p|X. But then
> since also p|Z, the
> >> equation Y^3=Z^3-X^3 implies p|Y^3, hence p|Y,
> contradicting the
> >> assumption that X,Y,Z are relatively prime.
> >>
> >
> >Great!
> >
> >Then
> >
> >(R,W) = 1 and R*W = Z^3 => R and W are perfect
> cubes.
> >
> >So R = z^3 and W = u^3 for some integers z and u.
> >
> >And
> >
> >Z^3 = R*W = z^3*u^3
> >=> Z = u*z
> >
> >Continuing with GG's proposed proof:
> >
> >> When Z is divisible by ,that is Z=u*3*z we see
> that
> >> W and R have as common
> >> divisor only 3
> >> Therfore W=(u^3)*3^(3-1) and R=3*z^n
> >
> >I assume this means "When Z is divisible by 3..."
> >
> >Anyone care to offer a proof of this next bit?
>
> Assume 3|Z.
>
> Then X,Y are not multiples of 3 since if 3 divides
> one of them, the
> equation X^3+Y^3=Z^3 together with 3|Z would imply 3
> divides the
> other, contrary to the assumption that X,Y,Z are
> relatively prime.
>
> Hence X*Y is not a multiple of 3.
>
> Now W*R=Z^3 => 3|(W*R) => 3|W or 3|R.
>
> Then the relation R=W^2-3*X*Y => R=W^2 mod 3 => both
> R,W are multiples
> of 3 (since we know at least one is).
>
> But 3|W => 9|W^2 => 9 does not divide R (since 9 does
> not divide
> 3*X*Y). Hence R is a multiple of 3 but not a multiple
> of 9.
>
> Then W*R=Z^3 => W*R is a multiple of 27 so W must be
> a multiple of 9.
>
> Since we know that in general (W,R)=1 or 3, in this
> case we must have
> (W,R)=3. Then the integers W/9,R/3 are relatively
> prime and since the
> product (W/9)*(R/3) is a perfect cube (=(Z/3)^3), it
> follows that R/3,
> W/9 are both perfect cubes.
>
> So one can write W=9*u^3, R=3*z^3 and then Z can be
> expressed as
> 3*u*z, as claimed.
>
> So this part of the proof is also correct.
>
> Without looking too carefully, it appears that in the
> next parts of
> the proof we are about to be dragged through a lot of
> elementary
> algebra -- substitutions, simplifications, etc, all
> of which is
> probably correct. This partly camouflages any later
> flaw, since one
> has to actually go through all the algebraic
> preliminaries to get at
> whatever is the key proof idea.
The key proof ideea is:
The last lines of the proof show that X^n +Y^n= Z^n implies that
Q^n+P^n=(Q+k*q*p+P)*[Q^(n-1) +g*q*p+P^(n-1)]=
(Q+P)*[Q^(n-1)+f*q*p+P^(n-1)]
>From development of Q^n+P^n in liniar factors according to
Lame method the above equality is Immpossible.
Therefore Fermat Last theorem is true.
THE END*
george
>
> quasi
.

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